Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Mathematics
(1/2.5)+ (1/5.8)+ (1/8.11)+.......upto n terms =
Q.
2.5
1
+
5.8
1
+
8.11
1
+.......upto
n
terms =
4514
239
KCET
KCET 2008
Sequences and Series
Report Error
A
4
n
+
6
n
7%
B
6
n
+
4
1
12%
C
6
n
+
4
n
74%
D
3
n
+
7
n
7%
Solution:
Let
S
n
=
2
⋅
5
1
+
5
⋅
8
1
+
8
⋅
11
1
+
…
+
(
3
n
−
1
)
(
3
n
+
2
)
1
=
3
1
[
2
1
−
5
1
+
5
1
−
8
1
+
………
+
3
n
−
1
1
−
3
n
+
2
1
]
=
3
1
[
2
1
−
3
n
+
2
1
]
=
6
n
+
4
n